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Basile [38]
3 years ago
10

Can you guys help me with this problem It is the picture

Mathematics
1 answer:
liq [111]3 years ago
8 0

Answer:

the last option

Step-by-step explanation:

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On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove
gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

v = \frac{d}{t}

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time t_1, distance of d_1, v = 40. So

v = \frac{d}{t}

40 = \frac{d_1}{t_1}

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is v = 60.

On the first day, he drove 2 more hours, which means that for the second day, the time is: t_1 - 2

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: d_1 - 20

Thus

v = \frac{d}{t}

60 = \frac{d_1 - 20}{t_1 - 2}

System of equations:

Now, from the two equations, a system of equations can be built. So

40 = \frac{d_1}{t_1}

60 = \frac{d_1 - 20}{t_1 - 2}

Find the total distance traveled in the two days:

We solve the system of equation for d_1, which gets the distance on the first day. The distance on the second day is d_2 = d_1 - 20, and the total distance is:

T = d_1 + d_2 = d_1 + d_1 - 20 = 2d_1 - 20

From the first equation:

d_1 = 40t_1

t_1 = \frac{d_1}{40}

Replacing in the second equation:

60 = \frac{d_1 - 20}{t_1 - 2}

d_1 - 20 = 60t_1 - 120

d_1 - 20 = 60\frac{d_1}{40} - 120

d_1 = \frac{3d_1}{2} - 100

d_1 - \frac{3d_1}{2} = -100

-\frac{d_1}{2} = -100

\frac{d_1}{2} = 100

d_1 = 200

Thus, the total distance is:

T = 2d_1 - 20 = 2(200) - 20 = 400 - 20 = 380

The total distance traveled in two days was of 380 miles.

For the relation between velocity, distance and time, you can take a look here: brainly.com/question/14307500

3 0
2 years ago
-4(2x + 12) in distributive property​
nikitadnepr [17]

Answer:

−8x − 48

Step-by-step explanation:

5 0
3 years ago
A solution of salt and water is 20% salt. If 25% of the water is removed, what is the percentage of salt in the new solution?
Ratling [72]
So lets say that this salt and water solution consists of 100 liters.

Then that means 20 liters is salt, and 80 liters out of the 100 is water.



Now, we remove 25% of the water. 25% of 80 liters is 20 liters. So, we are left with 80-20, or 60 liters.



Now in all we have 20 liters of salt, and 60 liters of water.

The percentage of salt in the new mixture will be 20/(20+60) = 20/80 = 25%.



Your answer is 25%
4 0
2 years ago
Complete the equation of the graphed linear function in point-slope form.
ozzi
Y = 2 x + 18 i think thats the answer :D
6 0
2 years ago
Cholesterol levels for a group of women aged 30-39 follow an approximately normal distribution with mean 190.14 milligrams per d
Anettt [7]

Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

2.

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!

8 0
3 years ago
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